Wednesday, 30 March 2016

[Q] Algebra Tricks - 3 [Approximation]

This is another useful article for you all and requires no prior knowledge of any sort.

Approximation is a very important tool that can help you solve some complex and time taking questions. I will solve the below questions from SSC CGL with approximation technique to give you an idea of how it works. But before that, some basic rules of approximation:

1. Establish a limit within which the variable is falling.
2. Neglect the smaller terms of the expression (fractions with Denominator>Numerator)
3. Please use this technique only when the options have a significant difference between them. E.g. If in a question the 4 options are A. 4, B. 5, C. 6, D. 7, you can't use the approximation technique because the options are fairly close.

These rules will make sense once you go through the below CGL questions-

Q. 1

Now how will you approach this question if you dont know how to solve it?
Given, x^4 + 1/x^4 = 119
We can safely assume that 3<x<4 because 3^4 = 81 and 4^4 = 256  (119 lies between 81 and 256). Moreover x will be closer to 3 as 119 is more close to 81 than 256
We have established the limit of the variable.
Let us take our first value. Go with x = 3.2
3.2^4 = 104 (approx), which is still a little away from 119
Hence let us take x = 3.3 as our second value
3.3^4 = 118 (approx) [PERFECT]
Now we have to find x^3 - 1/x^3
Note that 1/x^3 is negligible and hence we can neglect it
So just find the value of 3.3^3
Answer : (C)
Note : You won't take much time in calculating 3.2^4 or 3.3^4 if you know a fast method to calculate squares. I have written an article about it. You can check it here. 




Q. 2

Here again no need to figure out how to solve the question
√3 = 1.73, √5 = 2.23
Hence √x = 1.73 - 2.23 or x = 0.25
Put the value of x
(0.25)^2 - 16*0.25 + 6
= 0.0625 - 4 + 6
= 2 (approx)
Answer : (C)

Q. 3

x = √5 + 2 = 4.23
(x^4 - 1)x^2 = x^2 - 1/x^2
Neglect 1/x^2
x^2 = 4.23^2 = 17 (approx)
Answer : (A)


Q. 4

Again if you dont know how to solve the above question, then observe the above equation
If a=1, then LHS = 5.33 (which is little more than RHS, i.e., 5). We need to decrease the value of 'a'.
Hence let's take a = 0.9
5a + 1/3a = 4.8
Now LHS is more than RHS. We need to increase the value of 'a' slightly
So lets lock the final value a = 0.95 (Now no need to check the value of LHS for a=0.95)
9a^2 + 1/25a^2
Neglect 1/25a^2
9a^2 = 8 (approx)
Answer : (D)


Q. 5

x = 2 √3 = 2 + 1.73 = 3.73
Now you have to find the value of √x + 1/√x
√x = √3.73
You know that 19^2 = 361
Hence √3.73 = 1.9 (approx)
1/√x = 1/1.9 = 0.5
√x + 1/√x = 1.9 + 0.5 = 2.4 (which is close to √6)
Answer : (B)


Q. 6

3x - 1/4y = 6
Put x=1 and solve the equation for y
y = -1/12
Put x = 1 and y=-1/12 in the expression (4x - 1/3y)
You will get 8
Answer : (D)


Q. 7

Here again we can say that the approx value of x is 8, because 8^2 = 64
Put x = 8 in the expression
 = (64 - 1 + 16)/8
 = 10 (approx)
Answer : (A)

Q. 8)



Put x=2, the LHS becomes 6 and it is little more than RHS. So we need to decrease its value slightly
Let us take x = 1.8
LHS = (1.8)^2 + 1.8 = 5(approx)
LHS is almost equal to RHS, hence x=1.8 is a perfect value
Neglect 1/(x + 3)^3.
Now we only need to find the value of (x + 3)^3
(x + 3)^3 = (1.8 + 3)^3 = 110 (approx)
Answer : (A)


Below are some important formulas for Algebra, see if you can memorize them :)



If you have read CI-SI Tricks Part-1, I would request you to read the "annual instalments" section again. I have updated that topic...

Keep Reading :)

To buy the super-hit SSC Hack Book, follow the below link-

Monday, 28 March 2016

[Q] Simple and Compound Interest Tricks - 2


In this post I will give some extremely important formulas that will save your time. Read this post carefully and note down all the formulas in a piece of paper for quick revision.

Q. 1) If a sum of money becomes 3 times itself in 20 years at simple interest. What is the rate of interest?
In such questions apply the direct formula-
Rate of interest = [100*(Multiple factor - 1)]/T
So R = 100*(3 - 1)/20
Answer : 10%
Note : With this formula you can find Rate if Time is given and Time if rate is given.

Q. 2) 


In such questions, just write this line :
1st part : 2nd part : 3rd part = 1/(100+T1 * r) : 1/(100+T2 * r) : 1/(100+T3 * r)
 = 1/(100+2 * 5) : 1/(100+3 * 5) : 1/(100+4*5)
 = 1/110 : 1/115 : 1/120
 = 23*24 : 22*24 : 23*22
Hence 1st part = (23*24)/ (23*24 + 22*24 + 23*22) * 2379
Answer : 828
Note : Surprisingly, such questions when asked mostly have this same data, i.e., R=5% and T1, T2, T3 = 2, 3, 4 years, respectively. Only the Principal is changed. So it would be wise if you can just mug this line :
1st part : 2nd part : 3rd part = 23*24 : 22*24 : 23*22
Based on the above line, you would be able to solve such questions in a jiffy. 

But note that it will only work if the question is on Simple Interest. Like the below question appeared in SSC CGL Tier 2-
Q. 4
Here the data is samei.e., R=5% and T1, T2, T3 = 2, 3, 4 years, respectively. So we will write directly -
1st part : 2nd part : 3rd part = 23*24 : 22*24 : 23*22
A received(23*24)/ (23*24 + 22*24 + 23*22) * 7930
Answer : Rs. 2760

Q. 5) If a certain sum of money P lent out for a certain time T amounts to P1 at R1% per annum and to P2 at R2% per annum, then



The above formula is for calculating the Time, if the question asks the rate, then just interchange the rate and time. Hence the formula will become

R = (P1 - P2)*100/P2T1 - P1T2



Apply the formula:
R = (650-600)*100/600*6 - 650*4
R = 5%

Alternative Method :
You can solve such questions quickly without mugging the above formula. How?
The sum amounts to Rs. 600 in 4 years and Rs. 650 in 6 years. This means the simple interest is Rs. 50 for 2 years (because the amount increased from Rs. 600 to Rs. 650 in 2 years)
So the SI for 4 years is Rs. 100 (we have seen earlier than SI is proportional. So if SI = 100 for 2 years, then SI = 150 for 3 years, SI = 250 for 5 years and so on)

Now SI = Rs. 100; P = 600-100 = Rs. 500; t = 4 years
R = 100*SI/(P*t) = 10000/2000
Answer: 5%

For CI, the formula is different

Difference between CI and SI
This topic is very important from examination point of view. Note the following things-
If t=1 year, then SI = CI
If t=2 years then difference between CI and SI can be given by two formulas-


 
If t=3 years then difference between CI and SI can be given by two formulas-

In all the above formulas we have assumed that the interest is compounded annually

Let us solve some CGL questions

Q. 6

A = P(1+r/100)^t
Given, A=1.44P t = 2 years
1.44P = P(1 + r/100)^2
r = 20%
Answer : (D)

Q. 7

Here the interest is compounded half yearly, so the formulas we mugged earlier are of no use here. We will have to solve this question manually
SI = P*10*1.5/100 = 0.15P
CI = P(1 + 5/100)^3 - P = P(1.05^3 - 1)
Given CI - SI = 244
P(1.05^3 - 1) - 0.15P = 244
P = Rs. 32000
Answer : (C)

Q. 8

Time = 2 years
Hence apply the formula: Difference(D) = R*SI/200
CI - SI = R*SI/200
CI - SI = (12.5/200)*SI
510 = 1.0625*SI    [Since CI = Rs. 510]
SI = Rs. 480
Answer : (D)

       Q. 9

CI for 1st year = 10% of 1800 = Rs. 180
CI for 2nd years = 180 + 10% of 180 = Rs. 198
Total = 180+198 = Rs. 378
Hence time = 2 years
Or you can apply the formula
A = P(1+r/100)^t
Answer : (B)

Q. 10

2.5 = P*R*2/100 - P*r*2/100
2.5 = 10R - 10r
R - r = 0.25
Answer : (D)

       Q. 11

CI for 1st year = 5% of P = 0.05P
CI for 2nd year = 5% of P + 5% of (5% of P) = 0.05P + 0.0025P = 0.0525P
Total CI = 0.05P + 0.0525P = 0.1025P
Given, 0.1025P = 328
P = Rs. 3200
Answer : (C)
Note : You can solve this question by the formula A = P(1+r/100)^t as well

Q. 12

Note that in this question the CI for 2 years in not given, but the CI for the 2nd year is given.
CI for 2nd year = 10% of P + 10% of (10% of P) = 0.1P + 0.01P = 0.11P
Given, 0.11P = 132
P = Rs. 1200
Answer : (D)

Q. 13

Interest = Re. 1 per day = Rs. 365 for 1 year
SI = P*r*t/100
t=1, r=5%, SI = Rs. 365
So, P = 365*100/5 = Rs. 7300
Answer : (A)

Q. 14

We know 
Difference = P(r/100)^2(r/100 + 3)
P = Rs. 10000, r = 5%, t = 3 years
Hence D = Rs. 76.25
Answer : (C)

Q. 15

We know, R = [(y/x)^(1/T2 - T1) - 1]*100
 = [(1587/1200)^1/(3 - 1) - 1]*100
 = [(1587/1200)^1/2 - 1]*100
 = 3/20 * 100
 = 15%
Answer : (B)


So this is the end of CI and SI series. If you have any doubt in this topic, please drop a comment...

To buy the super-hit SSC Hack-Book, follow the below link
Buy SSC Hack-Book

Sunday, 27 March 2016

[Q] Simple and Compound Interest Tricks - 1

After reading this series (Part-1 and Part-2), you will be able to solve all the questions that are asked by SSC from this topic.

(1) The basic concept of CI and SI
Let's say you have Rs. 30000 and you keep this money in three different banks for 2 years(Rs. 10000 each). The three banks have different policy :
a) Bank A keeps your money at simple interest and offers you 5% interest
b) Bank B keeps your money at compound interest and offers you 5% interest. The interest is compounded annually.
Bank C keeps your money at compound interest and offers you 5% interest. The interest is compounded half-yearly.
After 2 years, which bank will give you most interest?
Let us calculate
Case (A)
Simple interest is calculated simply as (P*R*T)/100
Here P= 10000, r = 5% and T = 2 years
T = 2 years. Let's divide this period in two equal intervals of 1 year each
Hence SI received for the period 0 to 1 = 10000*5*1/100 = Rs. 500
SI received for the period 1 to 2 = 10000*5*1/100 = Rs. 500
So after 2 years, you will get Rs. 10000 + 500 + 500 = Rs. 11000
Note : Simple Interest is proportional. The interest received is same each year. So in the above example where SI was Rs. 500 for 1 years, that will mean the SI for 3 years is Rs. 1500, the SI for 5 years is Rs. 2500 and so on.

Case (B)
Compounded annually means whatever interest you will earn on first year, that interest will be added to the principal to calculate the interest for 2nd year. Let us see how
We know the CI formula is, Amount = P(1 + r/100)^t   (where Amount = P + CI)
CI received for the period 0 to 1 = Amount - Principal = 10000(1 + 5/100)^1 - 10000= Rs. 500
Now the amount received after 1 year will act as the Principal for calculating the Amount for next year
For calculating the amount for second year, you won't take P as 10000, but as Rs. 10500. So unlike SI where the interest was same each year, in CI the interest increases every year (because the principal increases every year)
CI received for the period 1 to 2 = Amount - Principal = 10500(1 + 5/100)^1 - 10500 = Rs. 525
Total interest received after two years = Rs. 500 + Rs. 525 = Rs. 1025
Total amount received after two years = Rs. 11025
Note :  In Case (b), to calculate the amount received after 2 years, I had divided the calculation into 2 intervals. It was done just for the sake of explanation. You can calculate the amount received after 2 years directly by 10000(1 + 5/100)^2

Case (C)
Just like case (b), where Principal was getting updated every year, in case (c) we will update the Principal every 6 months (half-year)
Since I have given the explanation in case (b), so in this case I will directly apply the formula
Amount received after 2 years = 10000(1 + 2.5/100)^4 = Rs. 11038 approx.

So sum it up
Case A - amount received after two years= Rs. 11000
Case B - amount received after two years= Rs. 11025
Case C - amount received after two years= Rs. 11038

Case C is giving the maximum return and rightly so because in Case (C) principal is increasing every 6 months.

Important formulas for Compund Interest - 



(2) A sum of money becomes x times in T years. In how many years will it become y times?

The approach to solve such questions is different for SI and CI

For SI : Formula = [(y - 1)/(x - 1)] * T
Q. 1) A sum of money becomes three times in 5 years. In how many years will the same sum become 6 times at the same rate of simple interest?
Solution : [(6 - 1)/(3 - 1)] * 5 = 5/2 * 5 = 12.5 years
Answer : 12.5 years

For CI : Formula = (logy/logx) * T
Now dont worry, I wont be asking you to study logarithms :)
But just remember one property of logs and that is enough to solve the questions
log(x y) = y.log(x)
Hence log(8) = log(23) = 3.log(2)
Q. 2) A sum of money kept at compound interest becomes three times in 3 years. In how many years will it be 9 times itself?
Solution : (log9/log3) * 3         ... (1)
log9 = log(32) = 2.log(3)
Put this value in (1)
= 2.log(3)/log(3) * 3
= 2 * 3 = 6 years
Answer : 6 years

(3) Interest for a number of days


Q. 3)

Here P = 306.25
R = 15/4 %
T = Number of days/365
Number of days = Count  the days from March 3rd to July 27th but omit the first day, i.e., 3rd March
   = 28 days(March) + 30 days(April) + 31 days(May) + 30 days(June) + 27 days(July)
   = 146 days
We know SI = (P * r * t)/100


Answer : Rs. 4.59

(4) Annual Instalments
This is the most dreaded topic of CI-SI. Before giving you the direct formula, I would like to tell you what actually is the concept of annual instalments(if you only want the formula and not the explanation, you can skip this part. But I would like you to read it)
Suppose you want to purchase an iPhone and its price is Rs. 100000 but you dont have Rs. 1 lakh as of now. What would you do? You have two options - either you can sell your kidney (which most the iphone buyers do :D), or you can go for instalments. But if you want to buy the iPhone through this instalment route, the seller will incur a loss. How? Had you paid Rs. 1 lakh in one go, the seller would have kept that money in his savings account and earned some interest on it. But you will pay this Rs. 1 lakh in instalments and that means the seller will get his Rs. 1 lakh after several years. So the seller is incurring a loss. The seller will compensate for this loss and will charge interest from you.
Let the annual instalment be Rs. x. and you pay it for 4 years.
After 1 year you will pay Rs. x and the seller will immediately put this money in his savings account (or somewhere else) to earn interest. He will earn interest on this Rs. x at the rate of r% for 3 years (because the total duration is 4 years and 1 year has already passed)
Hence the amount which the seller will get from this Rs. x instalment = x(1 + r/100)^3
After 2nd year, you will again pay Rs. x and the seller will earn interest on this Rs. x for 2 years.
The amount which the seller will get from this Rs. x instalment = x(1 + r/100)^2
After 3rd year, you will again pay Rs. x and the seller will earn interest on this Rs. x for 1 year.
The amount which the seller will get from this Rs. x instalment = x(1 + r/100)^1
After 4th year, you will pay Rs. x and your debt would be paid in full (no interest on this Rs. x)
The amount which the seller will get from this Rs. x instalment = x
Now let's add all the above four amounts to get the total amount the seller would get from all the instalments =
x(1 + r/100)^3 + x(1 + r/100)^2 + x(1 + r/100)^1 + x     ... (1)
Now, had you paid Rs. 1 lakh in one go (without going for the instalment route), then the amount received by the seller after 4 years would have been = 100000(1+r/100)^4      ... (2)
Now (1) should be equal to (2) because only then the two routes (instalment route and direct payment route) will give the same return and seller would have no problem in giving you the iPhone in instalments.

100000(1+r/100)^4 = x(1 + r/100)^3 + x(1 + r/100)^2 + x(1 + r/100)^1 + x  
[Remember the above equation for solving questions of compound interest]

P + P*r*4/100 = (x + x*r*3/100) + (x + x*r*2/100) + (x + x*r*1/100) + x  
[Remember the above equation for solving questions of simple interest]

Although for Simple Interest, we have a direct formula-

The annual instalment value is given by-

Now coming to the questions. There are two types of questions and they are bit confusing. In one type, the Amount is given and in another type, Principal is given

Type 1(Amount is given):

Q. 4) What annual installment will discharge a debt of Rs.6450 due in 4 years at 5% simple interest?
When the language the question is like "what annual payment will discharge a debt of ...", it means the Amount is given in the question.
In this question, the Amount(A) is given, i.e., Rs. 6450. So we can apply the formula directly
Here A = 6450, r = 5%, t = 4 years
Solution : 100*6450/[100*4 + 5*4*3/4]
Answer : 1500

Type 2 (Principal is given) :

Q. 5) A sum of Rs. 6450 is borrowed at 5% simple interest and is paid back in 4 equal annual installments. What is amount of each installment?

Here the sum is given. Sum means Principal.
But our formula requires Amount(A)
So we will calculate Amount from this Principal
A = P + SI = 6450 + 6450*5*4/100 = Rs. 7740
Now put the values in the formula
A = 7740, r = 5%, t = 4
Annual instalment = 100*7740/(100*4 + 5*4*3/2)
Answer : Rs. 1800

Q. 6) 

"Sum borrowed" means Principal.
This question is of Compound Interest and hence we cant apply the direct formula. We will solve this question with the help of the equation we derived earlier.

P(1+r/100)^2 = x(1+r/100) + x
P(1 + 5/100)^2 = 17640(1 + 5/100) + 17640
Solve for P, you will get P = Rs.32800
Answer : (B)

Q. 7) What annual instalment will discharge a loan of Rs. 66000, due in 3 years at 10% Compound Interest?

Solution : Here again the question is of "Compound Interest" and hence we will solve it by equation :
Let each annual instalment be of Rs. x. Note that in this question, amount is given
Amount = x(1 + 10/100)^2 + x(1 + 10/100)^1 + x
66000 = x (1.21 + 1.1 + 1)
So x = Rs. 19939.58

Q. 8) What annual instalment will discharge a loan of Rs. 66000, due in 3 years at 10% Simple Interest?
I have just converted Q.7 into Simple Interest
Now we can either solve it by direct formula, or by equation
By Equation method :
66000 = (x + x*10*2/100) + (x + x*10*1/100) + x
66000 = x(3 + 0.2 + 0.1)
x = Rs. 20000
By Direct formula method :
A = 66000, t = 3, r = 10%
x = 100A/[100t + t(t-1)r/2]
x = 100*66000/[100*3 + 3*2*10/2]
x = 6600000/(300 + 30)
x = Rs. 20000
Don't forget to read Part - 2

To buy the super-hit SSC Hack-Book, follow the below link-
Buy SSC Hack-Book

Saturday, 26 March 2016

[R] Reasoning Tricks - 3 [Counting squares, rectangles, triangles]

Many of you wanted shortcuts to count the number of triangles. So I am writing this post which will cover every possible scenario for counting squares, rectangles and triangles. If the question doesn't fall in any of the below category, you will have to count manually :)

1) Counting squares within a square

How many squares are there in the above figure?
Solution : There are 4 rows in the above figure. Hence n=4
Apply the formula [n(n+1)(2n+1)/6]
= 4*5*9/6
Answer : 30

2) Counting squares within a rectangle



How many squares are there in the above figure?
Solution : This rectangle is a 4 x 3 grid [4 rows and 3 columns]
So total number of squares is m(m+1)(2m+1)/6 + (n-m)*m(m+1)/2.where n is the larger dimension
Here n = 4, m = 3
Answer : 20

3) Counting rectangles within a rectangle



How many rectangles are there in the above figure?
Solution : For a 'n x m' grid ['n' is the number of rows and 'm' is the number of columns]
Number of rectangles = (m)(m+1)/2 (n)(n+1)/2 = m(m+1)(n)(n+1)/4
In the above figure
n = 4, m=6
Put the values
Answer : 210

4) Counting rectangles in a square
The formula to count the number of rectangles within a square can be obainted by putting n = m in the above formula (3rd case)
So, No. of rectangles = [n(n+1)/2]^2

Ques. How many rectangles are there in a chess board?

Solution : We know that the chess board is a 8 x 8 grid
Put n = 8 in the above formula
Answer : 1296

Note : From the above formulae, you only have to mug two, i.e., 2 and 3. Formula 1 can be obtained by putting n=m in formula 2 and formula 4 can be obtained by putting n=m in formula 3.

5) Counting triangles within a triangle



How many triangles are there in the above figure?

The formula is given by - 

You should note 3 things about this formula-
1. It is only applicable for matchstick arrangement (like the figure above)
2. n = number of unit triangles in a side. In this question n=4 (I have counted the triangles for you)
3. Those are not brackets enclosing the formula; they denote floor function. So after applying the formula, if you are not getting an integer, you should completely reject the decimal numbers. E.g. 17.12 will become 17 and 56.98 wil become 56
Caution : You don't have to round off the number, so 77.9 will be equal to 77 not 78.

Put n=4 in the formula
Answer : 27

Next question-

Here n=3
Apply the formula and you will get 13.125. Reject the decimal places-
Answer : 13

Keep Reading :

To buy the super-hit SSC Hack-Book, follow the below link-
Buy SSC Hack Book

Thursday, 24 March 2016

[R] Reasoning Tricks - 2 [Cubes/Dices]

In this article I will take up CUBES/DICES. They may pose a problem to you if you are not good at imagining things in 3D. So let's begin.
Firstly you should know that there are 2 types of dice - Standard and Ordinary.
In a standard dice, the sum of opposite faces is 7 [Remember that], so if they ask which face is opposite to 2, you can say it is 7-2 = 5. But mostly the questions are from ordinary dice, where the sum is not 7.

There are various rules to solve the problems on dice -

1. If two sides of Dices are common
If two sides of cubes are common( has same numbers or symbols), then the remaining two will be opposite to each other.


2. If one side of Dices is common
If one side of given dices is common then list these sides (numbers on them) either in clock-wise or anti-clockwise. Comparing the numbers obtained from both dices will give you the opposite numbers.

3. If one side is common and it's place is same in both dices.

If one side is common in both cubes and its place is also same, then the remaining two sides of respective dices will be the opposite to each other.

Now let's move to a typical CGL question - 

In this question the face opposite to 2 is asked, so just look at each cube and list the neighbours of 2
From 1st cube :
Neighbours of 2 = 3,4                  ...(1)
From 2nd cube :

Neighbours of 2 = 1,3                  ...(2)
In the third cube '2' is not present, so we will skip this cube
From 4th cube :

Neighbours of 2 = 4,6                   ...(3)
Now merge all the three results and prepare a consolidated list of the neighbours

Neighbours of 2 = 1, 3, 4, 6
Note the number which is missing from the list. Here it is 5
So 5 is opposite to 2.
Answer : 5

Note : For each face, there are 4 neighbours and 1 opposite face. So make sure when you make the final consolidated list for a face, it has 4 neighbours [like our list contained 4 neighbours]

Next question


From the above figure, neighbours of 6 = 1,3,5
Note that there are only 3 neighbours in the list and hence we can't be sure of the face that is opposite to 6. Either it can be 2 or 4 (both are missing from the list)
When you encounter such situations, pick the face which is featured in most of the cubes. We can see that 3 is present in three of the cubes, so let's pick it.
Neighbours of 3 = 1, 4, 5, 6
From above data we can deduce that the face opposite to 3 is 2. Now since 2 is opposite to 3, it cant be opposite to 6. Hence the face opposite to 6 is 4
Answer : 4

Unfolded Cubes
Remember the following things when you get an unfolded cube

In short, no matter which figure of an unfolded cube is given in the question, the alternate faces are opposite to each other.
Now a SSC question - 
From the unfolded cube it is clear that-
Circle is opposite to Triangle
Square is opposite to Heart
Diamond() is opposite to Clubs()
(A) In option A we can see that triangle is shown adjacent to circle, but we know that they are opposite. Hence option A is wrong
(C) In option C, square is adjacent to heart, but they are opposite. Hence C is wrong
(D) In option D, diamond is adjacent to clubs, but they are opposite. Hence D is wrong.
Answer : (B)


Keep Reading :

To buy the super-hit SSC-Hack Book, follow the below link
Buy SSC Hack-Book

Sunday, 20 March 2016

[Q] Geometry Tricks - 2 [Centres of a Triangle]

In this post I will share some very important formulae for Geometry. Geometry is all about theorems and properties and there are endless things to mug in it, but I will only mention those formulae that are asked by SSC every year and hence are indispensable. Make sure you absorb all these well.

Orthocentre, Incentre, Circumcentre and Centroid :

This topic is very important from SSC point of view. You all must be knowing the theory of these terms, but I will reiterate -
Orthocentre - Point of intersection of altitudes of a triangle
Incentre - Point of intersection of angle bisectors
Circumcentre - Point of intersection of perpendicular bisectors
Centroid - Point of intersection of medians
The above information, although important, is not sufficient enough to solve the questions. Each of these terms require different approach to solve the questions based on them.

Orthocentre












AHB + ACB = 180  [Whenever you see the word 'orthocentre' in the paper, you should immediately recall this formula]
Over the years SSC has asked many questions based on this simple formula. Like -
Q. 1

Answer : (A)

Incentre












∠AIC = 90 + ∠ABC/2 [Whenever you see the word 'incentre' in the paper, you should immediately recall this formula]

Circumcentre










While solving questions on circumcentre, you should imagine the above figure. Here two things are quite useful - 
∠AOB = 2∠ACB
OA = OB = OC [radius of the circle]. And hence,
∠OAB = ∠OBA
∠OCA = ∠OAC
∠OBC = ∠OCB
Q. 2

Each angle of an equilateral triangle is 60 and we have seen earlier that the angle at the centre is twice that of the triangle.
Hence AOC = 120
Answer : (C)

                                         Q. 3   

Answer : (D)

Centroid

Centroid divides the medians in the ratio 2:1. Hence for the median AE,
OA = 2/3AE
OE = 1/3AE
OA = 2OE
Similarly for the medians CD and BF.

Remember one more property related to centroid,
Area(ΔOBC) = 1/3Area(ΔABC)
Q. 4

We know DG = 1/2AG
Hence DG = 2cm
Answer : (A)

Some additional things to remember :

1) The orthocentre, incentre, circumcentre and centroid of an equilateral traingle coincide, i.e., a single point acts as all the centres.

2) For an equilateral triangle-

  • Inradius = h/3 = a/2√3
  • Circumradius or outer-radius= 2h/3 = a/√3
  • Height(h) = (√3/2)a
where a = side of the equilateral triangle

3) For a right-angled triangle
  • Orthocentre is at the right angle vertex
  • Circumcentre is the midpoint of the hypotenuse

Here is a question based on this fact-
Q. 5

Two angles are 45 degrees and hence the third angle is 90 degrees. The figure will look like this-

Radius of the circumcircle = 15, hence diameter = 30cm
AC = 30cm
We have to find AB and BC.
AB = BC [as ∠ACB = ∠ABC = 45]
AB = sin45 * AC = 30/√2 = 15√2
Answer : (C)


Interior and Exterior Angles
Another important topic of geometry is "Internal and External Angles" of polygon.
Now remember few formulae for such questions-
1) Sum of a interior angle and its corresponding exterior angle adds up to 180 degrees.
2) Sum of all exterior angles of a regular polygon is always 360.
3) Sum of interior angles = (n-2)*180
4) Each interior angle of a regular polygon is equal to [(n-2)*180]/n
5) Each exterior angle of a regular polygon is equal to 360/n

Mug all these formulae well and you will be able to solve all questions on interior and exterior angles easily.
Now let's solve some CGL questions
Q. 6
Given
(n - 2)*180 = 2*360
So n - 2 = 4
or n = 6
Answer : (B)
Q. 7

Note : Here the difference between the "angles" of two polygons is given. You might be wondering that they haven't mentioned the type of angle, whether interior or exterior. Well, it doesn't matter! Because as far as two polygons are concerned-
Difference between their interior angles = Difference between their exterior angles
But I will take the angles to be exterior, because that will make our calculations simple
Let the sides be 5x and 4x.
Measure of each exterior angle of Polygon 1 = 360/5x = 72/x
Measure of each exterior angle of Polygon 1 = 360/4x = 90/x
Given, 90/x - 72/x = 6
so x = 3
The two angles are 5x and 4x
5x = 5*3 = 15
4x = 4*3 = 12

Had I taken the angles to be interior - 
We have seen earlier that the measure of each angle of a polygon is [(n - 2)*180]/n.
Given [(5x - 2)*180]/5x - [(4x - 2)*180]/4x = 6
Solve it and you will get x = 3
So the angles are 15 and 12
Answer : (D)

Note : Please note that each and every formula that I have mentioned in this post is extremely important and I can guarantee that after mugging all these formulae you will be able to solve all the questions on orthocentre, incentre, circumcentre, centroid and interior-exterior angles...

Keep reading :)

To buy the super-hit SSC Hack Book, follow the below link-
Buy SSC Hack-Book